# Is the Schröder Equation valid for higher dimensional iterated maps?

The Schröder's functional equation is the eigenfunction equation for the composition operator given as:

$$\psi \circ y (x) = s \cdot \psi(x) ~~~~~~~~~~~ (1)$$

The interesting bit about it (at least for me) is related to the iteration of a function over a point. Suppose that $y,x \in \mathbb{R}$. If we denote $y(y(x)) = y \circ y(x) = y_2(x)$ we can extend the notation like so $y_n(x) = y \circ y_{n-1}(x)$ and, more generally $y_{n+m}(x) = y_n \circ y_m(x)$. It turns out that once a Schröder's conjugate $\psi(x)$ is found for $y(x)$, it is possible to compute the $n$-th composition of $y(x)$ straightforwardly like so:

$$y_n(x) = \psi^{-1}(s^n \cdot \psi(x))$$

By computing $y_n$ like this, not only the integer values of $n$ are well-defined, but also all the intermediate values, turning $n$ into a real parameter and the iterating function $y_n(x)$ into the Flow $y(n,x)$. Meaning that the entire orbit containing all iterates of $y$, including the fractional ones, is obtained. This equation is often solved for an arbitrary function $y(x)$ using the Carleman matrices to turn composition into multiplication.

Handy as it is, I could only find discussions about the Schröder equation, and its solution strategy, in the one-dimensional domain. So my question is whether there is a more general version of the Schröder's Equation for iterated maps of greater dimensionality, which generalizes equation (1) like so:

$$\vec \varphi \circ \vec u(\vec x) = s \cdot \vec \varphi(\vec x) ~~~ \forall \vec \varphi, \vec x, \vec u \in \mathbb{R}^n, s \in \mathbb{R}.$$

And if there is, whether or not we have a quasi-Carleman method to solve it as well.

Frankly, I have not seen many multivariate techniques for functional iteration. You might consider experimenting with powers, emulating Schroeder's technique. Take two variables, x and y, for simplicity.

Again for simplicity, consider diagonal vector functions, e.g., $$\vec u (x,y)= (x^2,y^3) ,$$ so $$\vec u_n(x,y)=(x^{2^n},y^{3^n}).$$ I guess $$\vec \varphi (\vec u (x,y)) = \operatorname{diag}(2,3) ~~\vec \varphi (x,y)$$ would do the trick for $\vec \varphi (x,y)=(\log x , \log y)$, so, by functional conjugacy, $$\vec u _n (x,y)= \vec \varphi^{-1}(\operatorname{diag}(2^n,3^n)~\vec \varphi(x,y) ).$$

Of course, for a scalar Schroeder eigenvalue as you have, you need the diagonal eigenvalue matrix to be proportional to the identity, so both entries 2 or 3 in our case. I took the slight generalization here to relax this restriction, even though there is no essential coupling between the two modes, so their orbits simply do not influence each other.

As for perturbative techniques around fixed points, given the suitable equation arrays, such are always available. But make no mistake, the above diagonal generalizations are trivial, i.e. they are simply addressing a multiplexing of the scalar equation of one variable, with innocuous parallel moves and no essential coupling. In point of fact, the renormalization group equation for multicoupling QFT systems is of this form, but people bypass this entire structure in favor of coupled ODE systems—not functional equations anymore. I can see you asking for more elaborate constructs next...

• One thing that caught my eye was the use of the "diag" symbol. The way I am understanding this is that you are working in the speific case where $s$ is a diagonal matrix. So for a multivariate Schroeder Equation I would have, in general, instead of a scalar eigenvalue $s$, an "eigenmatrix" $\bf S$ for which ${\vec u}_n = \vec \varphi^{-1} ( {\bf S}^n \cdot {\vec \varphi}(x,y) )$ ? In that case I could only find a continuous orbit for $\vec u$ if $\bf S$ is diagonalizable, right? So that I can apply fractional coefficients to it. – urquiza Aug 1 '18 at 21:07
• Yes. A diagonal "eigenmatrix". This is a simple tractable paradigm, and your appreciation of it is right, but it is ambitious to talk about what one "could only" do.... Imagination rules supreme... – Cosmas Zachos Aug 1 '18 at 21:10
• So is it right for me to conclude that the reason the proof for the one-dimensional Schröder's Equation validity does not extend to higher dimensions is because it stops being an eigenvalue equation and becomes this weird "eigenmatrix" equation? I can see how the method of playing around with examples can be illustrative o specific cases where it does apply, but I believe you're saying that no general method for solving $\vec \varphi \circ \vec u = {\bf S} \cdot \vec \varphi$ for $\vec \varphi$ with a known $\vec u$ exists. And apparently the very validity of the equation is not proven. Right? – urquiza Aug 1 '18 at 21:16
• "In point of fact, the renormalization group equation for multicoupling QFT systems is of this form, but people bypass this entire structure in favor of coupled ODE systems. I can see you asking for more elaborate constructs next..." Okay... It is probably a long shot but, is there a general way to go from this functional equation to a system of coupled ODEs? Or is this a particular feature of the QFT theory itself that allows for the students of renormalization group to make this switch? – urquiza Aug 1 '18 at 21:19
• You want a manual for a field that does not exist, properly. Diagonal eiegnmatrix problems are basically decoupled arrays of scalar problems, so all trivially tractable. The QFT RG techniques are very-very different than here, and basically coupled ODEs, not functional equations at the end of the day. This is an essentially different story, meriting different questions... – Cosmas Zachos Aug 1 '18 at 21:51